How to prevent derived class from making a private/protected virtual function public?

Question

There are good reasons for constructing the base class interface with all virtual functions as private or protected (see this). But then how does one prevent the derived cla

Answer 1

As Bjarne put it, the access control in C++ is meant to protect against Murphy, not Machiavelli. The same is true in general -- it's features are meant to protect against accidents, not people intentionally doing something wrong.

To an extent, using C++ means putting at least some degree of trust in the other people who will have access to your source code. If they want to badly enough, they can screw things up in all sorts of ways, and you can't do much of anything to stop them. If you want to place real restrictions on how your code is used, C++ is the wrong language for your purposes.

Edit: This isn't really an "argument" at all -- it's simply pointing out the basis upon which decisions were made. Since I have my copy of the D&E out from answering a previous question, I'll type a bit more if it in here¹:

It is more important to allow a useful feature than to prevent every misuse: You can write bad programs in any language. It is important to minimize the chance of accidental misuse of features, and much effort has been spent trying to ensure that the default behavior of C++ constructs is either sensible or leads to compile-time errors. For example by default all function argument types are checked -- even across separate compilation boundaries -- and by default all class members are private. However, a systems programming language cannot prevent a determined programmer from break the system so design effort is better expended on providing facilities for writing good programs than preventing the inevitable bad ones. In the longer run, programmers seem to learn. This is a variant of the old C "trust the programmer" slogan. The various type checking and access control rules exist to allow a class provider to state clearly what is expected from users, to protect against accidents. Those rules are not intended as protection against deliberate violation (§2.10).

In §2.10, he says, among other things:

The task of the protection system is to make sure that any such violation of the type system is explicit and to minimize the need for such violations.

Those goals appear to have been met here -- publicizing a protected base class member definitely requires explicit action in the derived class, and in 20+ years of writing C++ I can't remember ever needing (or even wanting) to do it.

¹§4.3, pgs. 115, 116.

Answer 2

it would make sense that the compiler would warn about such usage.

Why would it? It's up to each class's designer to decide about its external interface.

In C++, a base class has no special power to enforce derived class interface properties. The derived class could decide to make some overriding function private when the base function is public, or vice versa. The interface of the derived class is a contract with its clients, not with the base class (unless the base class is a client of the derived class, as with a curiously recurring template base class).

Answer 3

You could require a token argument which is only constructable by derived types. Of course, then they could just expose a subclass of the token. So you would have to give it a virtual destructor and RTTI check it.

protected:
class ProtectedToken { virtual ~ProtectedToken() { } };
virtual void my_tough_cookie(int arg,
  ProtectedToken const &tok = ProtectedToken() ) {
    assert ( typeid( tok ) == typeid( ProtectedToken ) );
    …
}

Of course this isn't a nice thing to do to anyone, including yourself.

Edit: Bah, it doesn't work. Even if it did, you could do public: using Base::ProtectedToken and defeat the protection that way. Another 15 minutes of my life wasted…

Answer 4

Access control in C++ possibly doesn't do what you want. It's not intended to enforce DRM-style constraints to stop you sharing your access. If A has access to B, then A can call B and use the result for any purpose, including returning it to another caller who doesn't have access to B.

The problem that's discussed in the article you link to isn't about A deliberately or maliciously sharing B. It's about what happens if you put a public virtual function in a published interface, and later try to change the class so that it uses their suggested Template Method patterns, including private virtual functions. Child classes have written public overrides of the virtual function, so you can no longer separate the two concerns (access and virtual-ness) without modifying all the child classes. The way I read it, the article does provide a solution to the problem it presents, and that solution is "never make virtual functions public in the first place".

Virtual functions should be treated very much like data members — make them private, until design needs indicate a less restricted approach is indicated. It is much easier to promote them to a more accessible level, than it is to demote them to a more private level.

The reason this doesn't solve your problem, is that they didn't consider your problem.

Answer 5

Promoting a private/protected virtual method to public in a derived class does not expose the base class method. It still cannot be called through a base class pointer. It does not become part of the interface of the base clase.

Answer 6

You can't. "virtualness" of a function and access type are two different unrelated concepts.